We can use recursion to solve this, traversing the tree DFS.
Thought process:
- Traverse the tree and increasing the depth each time
- If we found the leaf node, keep track of the maximum depth
- If we are already at the maximum depth, we increase our sum (because that is essentially the leaf node)
Link: Deepest Leaves Sum
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var deepestLeavesSum = function(root) {
let sum = 0;
let depth = 0;
let maxDepth = 0;
let traverse = (node, depth) => {
if(!node) return;
if(!node.left && !node.right){
if(depth > maxDepth){
sum = node.val;
maxDepth = depth;
}else if(depth === maxDepth){
sum += node.val;
}
}else{
traverse(node.left, depth+1)
traverse(node.right, depth+1)
}
}
traverse(root, 0)
return sum;
};